A piece of wire 24m long is to be used to form a square and or a rectangle whose length is three Times its width determine their maximum and minimum combined area

When it forms a square and a rectangle, Max area = 27 + 36 = 63 m^2

When it forms a square or a rectangle, min area = 27 m^2

Step-by-step explanation:

Since the whole length of the wire. =24 m

To form a square from it, all the four sides of the square should be the same, i.e

Each side of the square = 24/4

Each side of the square, l = 6 m

The area of the square = l^2

Area of the square = 6^2 = 36 m^2

For the rectangle, if the length = 3 * width

There are two equal lengths and two equal widths in a rectangle

2l = 3(2w)

Since the length is described to be 3 times the width, only 18 and 6 are three times each other that can be added to make 24.

I.e. 2l = 18, 2w = 6

l = 9 m, w = 3 m

The area of the rectangle = l*w

Area of the rectangle = 9*3 = 27 m^2

When it forms a square and a rectangle, Max area = 27 + 36 = 63 m^2

When it forms a square or a rectangle, min area = 27 m^2

6/20

step-by-step explanation:

solution:

we have to construct

∠m nt ≅ ∠p qr

the steps to draw this construction are

step 1: use a compass to draw an arc from point q which intersects the side pq at point a and the side qr at point b.

step 2: draw a segment nt and use the same width of the compass to draw an arc from point n which intersects the segment nt at a point x.

step 3:

step 4: join points n and y using a straightedge.

step 3:

option (d).maintaining the same width of the compass as ab, draw an arc from point x such that it intersects the arc drawn from n in a point y.

to draw   congruent angles , 1. width of compasses while drawing two angles should be same i.e from point at which angle has to be drawn, 2.the distance between the two points where the compass intersects the two sides should be same while drawing any two congruent angles.